What is the max frequency of sound in a given medium?

When talking about acoustic waves, medium continuity (pressure, density) is implicit. But can the wavelength of an acoustic wave be as small as intermolecular separation? For example, can ~10 THz sound waves travel in water?

• Yes, but they dissipate to heat in microns. May 18 '16 at 14:14
• I would guess the limit would be set by the number of molecules for a given volume of space. Using the experiment of a bell in a vacuum jar, one can determine the minimum number of molecules and their mean distance separation. If there aren't "enough" molecules, the frequency of the sound does not matter. If we are talking about "solids" and assuming that the sound energy does not break the molecular bonds, it should pass to the end of the solid, but possibly attenuated. May 25 '16 at 23:45
• "Inversely" Related: http://physics.stackexchange.com/q/121451/59023 and http://physics.stackexchange.com/q/192996/59023. Aug 25 '16 at 12:54
• There are limits for the minimum wavelength. These are related to the spacing between molecules, as you guessed. However, the maximum frequency is not necessarily related to this minimum frequency by $f=c/\lambda$. At very high frequencies the relationship may become non-linear. Regarding terahertz ultrasound, it is at the fore-front of research in the medical diagnostics by ultrasound. As people try to use it for medical diagnostics I suppose it propagates through water at least for a little bit.
– nasu
Apr 11 '19 at 16:43

I do not know how to answer your question directly. What happens at high (enough) frequencies is that the continuum model is not valid anymore. So, we can probably estimate that the maximum frequency is related to the minimum wavelength, that, is

$$\lambda_\min >> a$$

where $$a$$ is a characteristic scale at the micro-scale. Thus, when these two scales are far apart, we can trust the continuum model.

Below, you can see the derivation of the dispersion curves for a monoatomic crystal assuming that the interaction is with first neighbors and is mediated by linear springs. According to this model, there is a maximum limit for frequencies propagated through this material when the two length-scales are alike.

Example: Simple mass-spring lattice

Let us consider the system depicted in the following schematic Simple mass-spring lattice.

The force in the plane $$s$$ cause by the displacement of the plane $$s+p$$ is proportional to the difference $$u_{s+p}-u_s$$ of the displacements. We will consider only nearest-neighbor interactions, so $$p=\pm 1$$. The total force on $$s$$ comes from planes $$s=\pm 1$$: $$\begin{equation} F_s = c(u_{s+1}-u_s)+ c(u_{s-1}-u_s). \end{equation}$$ The constant $$c$$ is the stiffness between nearest-neighbor planes and will differ for longitudinal and transverse waves.

The equation of motion of the plane $$s$$ is $$m \ddot{u} = c(u_{s+1} + u_{s-1} -2 u_s),$$ assuming harmonic time dependence $$\exp(-i\omega t)$$ $$\begin{equation} -m\omega^2 u_s = c(u_{s+1} + u_{s-1} - 2u_s) \enspace . \tag{1} \end{equation}$$

Using the Bloch theorem $$u_{s\pm 1} = u_s e^{\pm i ka}.$$ So (1) is now $$-m \omega^2 u_s = c (u_s \exp(ika) + u_s \exp(-ika) - 2u_s)$$ and canceling $$u_s$$ from both sides, we have $$\omega^2 m = -c[ \exp(ika) + \exp(-ika) - 2 ] \enspace .$$

Using the identity $$2\cos ka = \exp(ika) + \exp(-ika)$$, and taking $$\Omega^2 = \omega^2/\omega_0^2 = \omega^2 m/c$$, we have the dispersion relation $$\begin{equation} \Omega^2 = 2 (1-\cos ka) \enspace . \tag{2} \end{equation}$$

The boundary of the first Brillouin zone lies at $$k=\pm \pi/a$$. We show from (2) that the slope of $$\Omega$$ versus $$ka$$ is zero at the zone boundary $$\frac{d\Omega^2}{d\, ka} = 2\sin ka=0$$ at $$ka=\pm \pi$$, $$\sin ka = 0$$.

By a trigonometric identity (2) may be written as $$\begin{equation} \Omega^2 = 4 \sin^2 \frac{1}{2}ka, \qquad \Omega = 2\left\vert \sin \frac{1}{2} ka\right\vert \enspace . \end{equation}$$ Plot of $$\Omega$$ versus $$ka$$. The region of $$ka<<1$$ or $$\lambda/a>>1$$ corresponds to the continuum approximation; where $$\Omega$$ is directly proportional to $$ka$$ (and is enclosed between the dashed lines). The First Brillouin Zone is placed between $$-1$$ and $$1$$.

• So what is your conclusion? How is this related to the an answer to the OP question?
– nasu
Apr 11 '19 at 16:36
• @nasu, that you can't have a wave that is shorter than the intramolecular distance Apr 11 '19 at 16:52
• Is this conclusion a result or your calculations? Or it's an assumption you use?
– nasu
Apr 11 '19 at 17:23
• @nasu, a result. Apr 11 '19 at 17:32